When calculating a simple diagram I came across an ambiguity in the conservation of momentum, i.e. it seems to me that the particle could come out of the process with opposite momentum with respect to its initial state. This might be a complete triviality, in that case I apologize, but I can't find my mistake. Take the following Feynman diagram in λϕ4 theory
Call x and y the two external points, z1 and z2 the two internal points. Using Feynman's rules in position space this is proportional to the integral
∫dz1dz2DF(x−z1)DF(z1−z2)2DF(z2−z2)DF(z1−y)
Where DF is the Feynman propagator of the Klein Gordon field
DF(x−y)=∫d4p(2π4)ip2−m2+iεe−ip(x−y)
Inserting this into the diagram, and writing dq=d4px(2π)4d4py(2π)4d4p1(2π)4d4p2(2π)4d4p3(2π)4 for the sake of brevity
∫dz1dz2DF(x−z1)DF(z1−z2)2DF(z2−z2)DF(z2−y)==∫dz1dz2∫dqe−ipx(x−z1)p2x−m2+iεe−ip1(z1−z2)p21−m2+iεe−ip2(z1−z2)p22−m2+iε1p23−m2+iεe−ipy(z1−y)p2y−m2+iε==∫dz1dz2∫dqe−iz1(−px+p1+p2+py)p2x−m2+iεe−iz2(−p1−p2)p21−m2+iεe−i(pxx−pyy)p22−m2+iε1p23−m2+iε1p2y−m2+iε
The momenta in the exponentials are gonna give δ(4)(px−p1−p2−py)δ(4)(p1+p2)
Which implies conservation of momentum, and everyone is happy. But let's get back to the first expression of the integral, where I wrote
∫dz1dz2DF(x−z1)DF(z1−z2)2DF(z2−z2)DF(z1−y)
I could have written
∫dz1dz2DF(x−z1)DF(z1−z2)2DF(z2−z2)DF(y−z1)
because the propagator is symmetric. This seemingly harmless change changes the exponential that yields the delta in
e−iz1(−px+p1+p2−py)
and the conservation of momentum will be δ(4)(px−p1−p2+py)δ(4)(p1+p2)
which in the end implies px=−py, or, the particle comes out with opposite momentum, which seems weird to me. Have I made a mistake or is the sign of the momentum not important in some way?
Answer
Let's call the OP's diagram Σ(x−y), which can be rewritten as Σ(x−y)=A∫dpxdpyδ(px−py)e−i(pxx−pyy)DF(px)DF(py),
Playing the game of the second part of the question (i.e. replacing DF(z1−y) by DF(y−z1)), we would end up with Σ(x−y)=A∫dpxdpyδ(px+py)e−i(pxx+pyy)DF(px)DF(py),
What confuses the OP is the question of momentum conservation, which looks strange in the second equation. This is because we should look at momentum conservation in momentum space (whereas we are still in real space here), i.e. we should compute Σ(p1,p2)=∫dxdyeip1x−ip2yΣ(x−y).
Starting from either expression, we obtain Σ(p1,p2)=δ(p1−p2)ADF(p1)DF(p2),
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